如何在查询生成器laravel中执行此SQL
如何在查询生成器laravel中执行此SQL如何在查询生成器laravel中执行此SQL,sql,laravel,query-builder,laravel-query-builder,Sql,Laravel,Query Builder,Laravel Query Builder,如何在查询生成器laravel中执行此SQL select Doctor_id from doctors where Doctor_id NOT IN (SELECT Doctor_id from report_reviewers WHERE Report_id = 26 ) 试试这个 $result = DB::table('doctors') ->whereNotIn('doctor_id', function($q){ $q->from('report_review
select Doctor_id from doctors where Doctor_id NOT IN
(SELECT Doctor_id from report_reviewers WHERE Report_id = 26 )
试试这个
$result = DB::table('doctors')
->whereNotIn('doctor_id', function($q){
$q->from('report_reviewers')
->select('Doctor_id')
->where('Report_id', '=', 26)
})
->select('doctor_id')
->get();
当你有疑问的时候,如果你知道原始SQL,你可以简单的做
$query = 'select Doctor_id from doctors ...';
$result = DB::select($query);
而且,所有的功劳都归于这位天才
更新
缺少的参数来自您正在使用的闭包
->whereNotIn('doctor_id', function($q,$id){
$q->from('report_reviewers')
->select('Doctor_id')
->where('Report_id',$id);
})
您需要像这样传递$id
变量
->whereNotIn('doctor_id', function($q) use ($id) {
$q->from('report_reviewers')
->select('Doctor_id')
->where('Report_id',$id);
})
请重试并查看。这是怎么回事:$result=DB::table('doctors')->join('users','doctors.UserID','=','users.id')->where('doctors.Acceptable',1)->whereNotIn('doctor\u id',function($q,$id){$q->from('report\u reviewers')->选择('doctor\u id'))->where('Report_id',$id);})->select('doctors.doctor_id','users.name')->get();在论点2($id)中丢失了哦,没关系。你忘恩负义。我的密码成功了,你接受了我的答案。现在你需要进一步的帮助,你不接受我的回答,即使它有效。一切都很好,祝你有一个愉快的一天。感谢EddyTheDove帮助我编辑此$this->xx=$id$result=DB::table('doctors')->join('users','doctors.UserID','=','users.id')->where('doctors.Acceptable',1)->whereNotIn('doctors_id',function($q){$q->from('report_reviewers')->select('doctors_id')->where('report_id',$this->xx);)->select('documents.doctor_id','users.name')->get();